a cyclist start from rest and accelerate uniformly to achieve a velocity of 12m/sec in 5 sec.immediately he applied brake and come to rest in next 4 sec .calculate
a)distance traveled during acceleration
b)distance traveled during retardation.
c)total displacement during whole journey.
a) (average velocity)x(5s) = ___
b) (average velocity)x(4s) = ___
c) Add a) and b)
To find the distance traveled during acceleration, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = distance traveled
u = initial velocity (0 m/s since the cyclist starts from rest)
a = acceleration
t = time (5 seconds in this case)
a) Distance traveled during acceleration:
Since the cyclist starts from rest, the initial velocity u is 0 m/s. The final velocity v is given as 12 m/s, and the time taken t is 5 seconds.
Using the equation s = ut + (1/2)at^2, we can rearrange it to solve for the distance s:
s = (1/2)at^2
Substituting the values given:
s = (1/2) * a * (5^2)
s = 25a
Therefore, the distance traveled during acceleration is 25 times the acceleration.
To find the distance traveled during retardation, we can use the same kinematic equation:
b) Distance traveled during retardation:
Since the cyclist comes to rest, the final velocity v is 0 m/s. The initial velocity u is given as 12 m/s, and the time taken t is 4 seconds.
Using the equation s = ut + (1/2)at^2, we can rearrange it to solve for the distance s:
s = ut + (1/2)at^2
Substituting the values given:
s = 12 * 4 + (1/2) * a * (4^2)
s = 48 + 8a
Therefore, the distance traveled during retardation is 48 + 8 times the acceleration.
c) Total displacement during the entire journey:
Displacement is the difference between the final position and the initial position.
Since the cyclist starts from rest and returns to rest, the initial and final positions are the same. Therefore, the displacement during the whole journey is 0.
To summarize:
a) Distance traveled during acceleration = 25a
b) Distance traveled during retardation = 48 + 8a
c) Total displacement during the whole journey = 0